10 Asymmetric Barrier Presentation...Asymmetric Barrier Design 2/16/2017 12 Reinforcing Design Barrier reinforcing shall be designed in accordance with section A13.3 of the LRFD Specifications

Asymmetric Barrier Design 2/16/2017

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Asymmetric Barriers

Pete White, PESystems Assessment Manager - Greenfield, INDOT

February 16, 2017

Overview What is an Asymmetric Barrier?

Median barrier with unbalanced roadway elevations

Unbalance


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Overview When Do We Need to Design?

Unbalance < 2’, provide equivalent overturning resistance as standard unreinforced median barrier

Unbalance < 2’

Overview When Do We Need to Design?

Unbalance > 2’, design as a reinforced retaining wall in accordance with AASHTO LRFD Bridge Design Specifications

Unbalance > 2’


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Overview What Shape is the Barrier?

Based on standard 33” or 45” median barrier (E 602-CCMB-04)


Geometry of faces should remain standard

Standard face geometry

Standard face geometry

Width may be increased

Vertical face


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Joint ‘B’ spacing doesn’t apply to reinforced (i.e. > unbalance) barriers (E 602-CCMB-02)

Overview How can Resistance be Increased?

Increase width and/or embedment

Increased width will increase barrier weight; not efficient for higher loads

Increased embedment will increase passive resistance; has practical limits


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Add a spread footing

Spread footing increases overturning resistance by adding width and increases sliding resistance by adding weight, but increases construction complexity and excavation widths


Incorporate mechanically stabilized earth (MSE) (reduces resistance demand)

MSE reduces lateral earth pressures, but increases construction complexity and excavation width

Wire Face Wall


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Load Cases Case 1 – During Construction

Lower pavement not included in analysis, regardless of anticipated construction sequence

Omit lower pavement during construction to account for changes in MOT and future pavement replacement

Load Cases Case 1 – During Construction

Vehicle collision force (CT) not required if temporary barriers utilized

CTTemporary barriers shall be shown in the plans


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Load Cases Case 2 – Final In-Service Configuration

Lower pavement in place and vehicle collision force applied

Lower pavement in place and providing passive resistance

CT

Vehicle collision force shall be applied to the top of the barrier, in addition to all other static loads

Loading Driving Forces

Active earth pressure (EHa), live load surcharge (LS), dead load surcharge (ESa), and vehicle collision force (CT); others as applicable

CT

ESaLSEHa

Water pressure may be omitted of adequate drainage is specified


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Vehicle collision force (CT) varies depending on the element being designed

CT = 10 kips, for stability analysisCT = LRFD section 13, for barrier reinforcing

NCHRP Report 663 indicates that a vehicle collision force of 10 kips is appropriate for static equilibrium (overturning and sliding) analysis. Forces given in LRFD section 13 are impact loads appropriate for reinforced concrete design.

NCHRP Report 663 indicates that a vehicle collision force of 10 kips is appropriate for static equilibrium (overturning and sliding) analysis. Forces given in LRFD section 13 are impact loads appropriate for reinforced concrete design.


Load combinations and load factors shall be as given in LRFD Table 3.4.1.1

Load factors during construction may be reduced as appropriate, per section 3.4.2 of the LRFD Specifications


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Loading Resisting Forces

Passive earth pressure (Rep), dead load surcharge (ESp), pavement passive resistance (EHpp), barrier self-weight (DLb), and sliding resistance (R)

DLb

Rep ESp EHpp

R

Sliding and passive may be used simultaneously, provided appropriate resistance factors are used (LRFD 10.5.5.2.2-1)

Passive resistance of lower pavement shall not exceed allowable compressive strength

Stability Analysis Sliding Check

Sliding shall be checked per Section 10.6.3.4 of the LRFD Specifications


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Stability Analysis Overturning Check

Overturning shall be checked per Section 11.6.3.3 of the LRFD Specifications

Stability Analysis Bearing Resistance Check

Bearing resistance shall be checked per Section 11.6.3.2 of the LRFD Specifications

B’

qapplied

DLb


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Stability Analysis Design Height and Length

Design height of the barrier shall not be taken as less than the average height of barrier within a 100 foot length of barrier, or the length of barrier between non-load transferring joints in the barrier, whichever is less

100’ Max.

Hmax

Hmin

Hdesign = (Hmax – Hmin) x ½

However,

(Hmax – Hdesign) ≤ 1 foot

Hdesign = (Hmax – Hmin) x ½

However,

(Hmax – Hdesign) ≤ 1 foot

Stability Analysis Design Height and Length

If at least 25 feet of the barrier within this length has an unbalanced height > 2 feet, the barrier should be designed for an unbalanced height > 2 feet

100’ Max.

Design as a reinforced retaining wall in accordance with AASHTO LRFD Bridge Design Specifications

Design as a reinforced retaining wall in accordance with AASHTO LRFD Bridge Design Specifications

25’

H = 2’

H > 2’


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Reinforcing Design Barrier reinforcing shall be designed in accordance with

section A13.3 of the LRFD Specifications

CT (LRFD Table A13.2.1,

TL-4 = 54 kips,TL-5 = 124 kips)

H

Lateral earth pressures not required in conjunction with vehicle collision force

Example – Less than 2’ Calculate the overturning resistance of standard

median barrier

Pstd

H

DLb

B/2

Assumed point of rotation

Pstd = (DLb x B/2) / HPstd = (DLb x B/2) / H


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Example – Less than 2’ Calculate the overturning moments due to earth

pressures and convert that to and equivalent force at the top of the barrier

Hub < 2’

LSEHa

Peq

H

Mot = (EHa x Hub/3) + (LS x Hub/2)

Peq = Mot/H

Mot = (EHa x Hub/3) + (LS x Hub/2)

Peq = Mot/H

Notes:1. Lateral earth pressure

forces should be factored

2. Upper pavement has been conservatively assumed as soil in this example

Notes:1. Lateral earth pressure

forces should be factored

2. Upper pavement has been conservatively assumed as soil in this example

Example – Less than 2’ Calculate the total required lateral moment

resistance and determine the required barrier weight

Preq = Pstd + Peq

H

Mreq = Preq x H

WTreq = Mreq/(B/2)

Mreq = Preq x H

WTreq = Mreq/(B/2)

Note:Moment resistance can

also be increased by widening the barrier instead of, or in conjunction with, increasing the barrier depth

Note:Moment resistance can

also be increased by widening the barrier instead of, or in conjunction with, increasing the barrier depth

WTreq

B/2


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Example – Greater than 2’ Case 1 – During Construction

Assume an embedment depth and calculate the driving lateral forces and overturning moment

ESaLSEHa

Assume an embedment depth

Notes:1. Temporary barrier used instead of collision force2. Drainage provided instead of water pressure

Notes:1. Temporary barrier used instead of collision force2. Drainage provided instead of water pressure

d1/3

d1/2

Fdriving = ∑(i x Fi), units force/length

Moverturning = ∑(i x Fi x di), units force x length/length

Fdriving = ∑(i x Fi), units force/length

Moverturning = ∑(i x Fi x di), units force x length/length


Calculate the resisting forces and moments

DLb

RepR

d1/3

R DLb x tan()

Fresisting = (R x R) + (Rep x Rep)

Mresisting = (Rep x Rep x d1/3)

R DLb x tan()

Fresisting = (R x R) + (Rep x Rep)

Mresisting = (Rep x Rep x d1/3)


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Check sliding resistance

If, Fresisting > Fdriving, Sliding resistance is adequate

If, Fresisting < Fdriving, Sliding resistance is deficient. Increase resistance by increasing barrier weight (width), embedment depth, or decrease driving force (MSE, etc.)


If, Fresisting < Fdriving, Sliding resistance is deficient. Increase resistance by increasing barrier weight (width), embedment depth, or decrease driving force (MSE, etc.)

FresistingFdriving


Check overturning Determine the net factored lateral load moments

Mnet = Moverturning – Mresisting (moments taken about center of barrier, B/2) Mnet = Moverturning – Mresisting (moments taken about center of barrier, B/2)

MresistingMoverturning

DLb

B


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Check overturning Calculate the eccentricity of the applied loads and check

against the maximum allowable eccentricity

emax = B/3 (per LRFD section 11.6.3.3)

e = Mnet/DLb

If emax > e, overturning resistance is acceptable

If emax < e, overturning resistance is deficient, increase resistance by increasing barrier weight (width), embedment depth, or decrease driving force (MSE, etc.)


e = Mnet/DLb


If emax < e, overturning resistance is deficient, increase resistance by increasing barrier weight (width), embedment depth, or decrease driving force (MSE, etc.)

Mnet

DLb

eB


Check bearing pressure Calculate the effective barrier width due to the eccentricity of

the applied loads and determine the applied bearing pressure

B’ = B – 2e

qapplied = DLb/ B’

qmax allowable = Per geotechnical recommendations

If qmax allowable > qapplied, bearing pressure is acceptable

If qmax allowable < qapplied, bearing pressure is deficient. Increase resistance by increasing barrier weight (width), embedment depth, or

decrease driving force (MSE, etc.)

B’ = B – 2e




If qmax allowable < qapplied, bearing pressure is deficient. Increase resistance by increasing barrier weight (width), embedment depth, or

decrease driving force (MSE, etc.)

DLb

B’

qapplied


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Example – Greater than 2’ Case 2 – Final Condition

Using the Case 1 section geometry, calculate the driving forces and overturning moment applied over the design length of the barrier

ESaLSEHa Embedment depth from Case 1

d1/3

d1/2

Fdriving = ∑(i x Fi), units force

Moverturning = ∑(i x Fi x di), units force x length

Fdriving = ∑(i x Fi), units force

Moverturning = ∑(i x Fi x di), units force x length

CT (10 kips)


Calculate the resisting forces and moments

DLb

RepR

d1/3

R DLb x tan()

Fresisting = ∑(i x Fi)

Mresisting = ∑(i x Fi x di)

R DLb x tan()

Fresisting = ∑(i x Fi)

Mresisting = ∑(i x Fi x di)

ESp

EHpp

d1/2 d1/3

Note:The passive resistance of the

lower pavement shall not exceed the allowable compressive strength

Note:The passive resistance of the

lower pavement shall not exceed the allowable compressive strength


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Check sliding resistance


If, Fresisting < Fdriving, Sliding resistance is deficient. Increase resistance by increasing barrier weight (width), embedment depth, barrier design length or decrease driving force (MSE, etc.)


If, Fresisting < Fdriving, Sliding resistance is deficient. Increase resistance by increasing barrier weight (width), embedment depth, barrier design length or decrease driving force (MSE, etc.)

FresistingFdriving


Check overturning Determine the net factored lateral load moments

Mnet = Moverturning – Mresisting (moments taken about center of barrier, B/2) Mnet = Moverturning – Mresisting (moments taken about center of barrier, B/2)

MresistingMoverturning

DLb

B


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Check overturning Calculate the eccentricity of the applied loads and check

against the maximum allowable eccentricity


e = Mnet/DLb


If emax < e, overturning resistance is deficient. Increase resistance by increasing barrier

weight (width), embedment depth, barrier design length or decrease driving force (MSE,

etc.)


e = Mnet/DLb


If emax < e, overturning resistance is deficient. Increase resistance by increasing barrier

weight (width), embedment depth, barrier design length or decrease driving force (MSE,

etc.)

Mnet

DLb

eB


Check bearing pressure Calculate the effective barrier width due to the eccentricity of

the applied loads and determine the applied bearing pressure

B’ = B – 2e




If qmax allowable < qapplied, bearing pressure is deficient. Increase resistance by increasing barrier weight (width), embedment depth, design length or decrease driving force (MSE, etc.)

B’ = B – 2e




If qmax allowable < qapplied, bearing pressure is deficient. Increase resistance by increasing barrier weight (width), embedment depth, design length or decrease driving force (MSE, etc.)

DLb

B’

qapplied

Documents

10 Asymmetric Barrier Presentation...Asymmetric Barrier Design 2/16/2017 12 Reinforcing Design Barrier reinforcing shall be designed in accordance with section A13.3 of the LRFD Specifications